The specific heat constant for iron is: [tex]c_{Fe} = 470 \frac{J}{kg \cdot K}[/tex]
Since the unit of measurement for the specific heat constant is Joules per kilogram Kelvin, we transform the temperatures from degrees Celsius to degrees Kelvin: T1 = 12 °C = 273.15 + 12 = 285.15 K T2 = 84 °C = 273.15 + 84 = 357.15 K
We also convert the mass from grams to kilograms: m = 46 g = 0.046 kg
The amount of heat absorbed by the skillet is: [tex]Q = m \cdot c_{Fe} \cdot \Delta T = m \cdot c_{Fe} \cdot (T_2 - T_1)[/tex]
Replacing the above values, we get: [tex]Q = 0.046 \cdot 470 \cdot (357.15 - 285.15) = 1557[/tex]
The answer is 1557 J (joules) or 1.557 kJ (kilojoules).
If the material is aluminium, use the specific heat of aluminium in your calculation, which is: [tex]c_{Al} = 920 \frac{J}{kg \cdot K}[/tex]
